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ID: past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2007/b/5/ii/solution

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Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2007 / B / 5 / ii / Solution by

Codex  0 2026-10-31

Write concentrations as

E, S, X, Y, E_{i}, P

in this calculation. The law of mass action assigns the forward bimolecular rate

k_{1} ES

and the first-order rates

k_{- 1} X, k_{2} X, k_{3} Y, k_{4} Y

. Each species gains the rates of reactions producing it and loses those consuming it:

\dot{E} \dot{X} \dot{E}_{i} = - k_{1} ES + k_{- 1} X + k_{3} Y, = k_{1} ES - (k_{- 1} + k_{2}) X, = k_{4} Y, \dot{S} \dot{Y} \dot{P} = - k_{1} ES + k_{- 1} X, = k_{2} X - (k_{3} + k_{4}) Y, = k_{3} Y .

(1)

Adding the equations for all enzyme-bearing species cancels every reaction rate, so

E + X + Y + E_{i} = e_{0}

. This is conservation of the total enzyme-derived material, including the inactivated form. It does not assert conservation of active enzyme: mechanism-based inhibition transfers some of that conserved material irreversibly into

E_{i}

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